Right circular cone P is similar to right circular cone Q, and cone P is smaller than cone Q. The...

1. TRANSLATE the problem information

• Given information:
  • Two similar right circular cones P and Q (P is smaller)
  • Volume of P = 160 cubic meters
  • Volume of Q = 540 cubic meters
  • Surface area of Q = 180 square meters
• Find: Difference between surface areas of Q and P

2. INFER the scaling relationships

• Since the cones are similar, there's a constant ratio k between corresponding linear dimensions (P to Q)
• This means: Volume ratio = \(\mathrm{k^3}\) and Surface area ratio = \(\mathrm{k^2}\)
• Strategy: Use the known volumes to find k, then use k to find the unknown surface area

3. SIMPLIFY the volume ratio

• Volume(P)/Volume(Q) = \(\frac{160}{540}\)
• Simplify by dividing both by 20: \(160÷20 = 8\), \(540÷20 = 27\)
• So the ratio is \(\frac{8}{27}\), which equals \(\mathrm{k^3}\)

4. SIMPLIFY to find the linear scale factor k

• \(\mathrm{k^3 = \frac{8}{27}}\)
• Take cube root: \(\mathrm{k = \sqrt[3]{\frac{8}{27}} = \frac{\sqrt[3]{8}}{\sqrt[3]{27}} = \frac{2}{3}}\)
• (Note: \(\mathrm{8 = 2^3}\) and \(\mathrm{27 = 3^3}\), so this works out perfectly)

5. INFER the surface area relationship

• Surface area ratio = \(\mathrm{k^2 = (\frac{2}{3})^2 = \frac{4}{9}}\)
• This means Surface Area(P)/Surface Area(Q) = \(\frac{4}{9}\)

6. SIMPLIFY to find surface area of cone P

• Surface Area(P)/180 = \(\frac{4}{9}\)
• Surface Area(P) = \(180 × \frac{4}{9}\)
  \(= 180 × 4 ÷ 9\)
  \(= 720 ÷ 9\)
  \(= 80\) square meters

7. Calculate the final answer

• Difference = Surface Area(Q) - Surface Area(P) = \(180 - 80 = 100\)

Answer: 100 square meters

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Not recognizing the connection between volume scaling (\(\mathrm{k^3}\)) and surface area scaling (\(\mathrm{k^2}\)) for similar figures. Students might try to work directly with the volume and surface area formulas for cones, getting bogged down in unnecessary complexity with radius and height calculations. This leads to confusion and guessing rather than systematic solution.

Second Most Common Error:

Poor SIMPLIFY execution: Making calculation errors when simplifying \(\frac{160}{540}\) or when taking the cube root of \(\frac{8}{27}\). For example, incorrectly simplifying the fraction to something like \(\frac{16}{27}\) instead of \(\frac{8}{27}\), or not recognizing that \(\sqrt[3]{\frac{8}{27}} = \frac{2}{3}\). These errors cascade through the solution, leading to wrong surface area calculations and final answers.

The Bottom Line:

This problem tests whether students can recognize and apply scaling relationships for similar 3D figures rather than getting caught up in complex geometric formulas. The key insight is that similar figures scale predictably: linear dimensions by k, areas by \(\mathrm{k^2}\), and volumes by \(\mathrm{k^3}\).